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What is the minimal cardinal of a family which shatters all d-subsets of a finite set?
N Chevallier, A Fruchard
To cite this version:
N Chevallier, A Fruchard. What is the minimal cardinal of a family which shatters all d-subsets of a finite set?. 2015. �hal-01194700�
What is the minimal cardinal of a family which shatters all d-subsets of a finite set?
N. Chevallier and A. Fruchard September 7, 2015
In this note, d ≤ n are positive integers. Let S be a finite set of cardinal |S| = n and let 2S denote its power set, i.e. the set of its subsets. A d-subset of S is a subset of S of cardinal d.
LetF ⊆2S and A ⊆S. The trace of F on A is the family FA ={E∩A ; E ∈ F }. One says that F shattersA if FA= 2A. The VC-dimension ofF is the maximal cardinal of a subset of S that is shattered by F [7]. The following is well-known [7, 4, 5]:
Theorem 1. (Vapnik-Chervonenkis, Sauer, Shelah)
If VC-dim(F)≤d (i.e. if F shatters no (d+ 1)-subset of S) then |F | ≤ c(d, n), where c(d, n) = n0
+· · ·+ nd . Moreover this bound is tight: It is achieved e.g. for F = ≤Sd
, the family of all k-subsets of S, 0≤k ≤d.
A first natural question is:
Question 1 . Assume a family F ⊆ 2S is maximal for the inclusion among all families of VC-dimension at most d. DoesF always have the maximal possible cardinal c(d, n)?
Let us define the index of F as follows:
IndF = max{d∈ {0, ..., n} ; F shatters all d-subsets ofS}.
Let C(d, n) = min{|F | ; IndF = d}. For instance, we have C(1, n) = 2, with the (only possible) choice F ={∅, S}. Of course we have 2d≤C(d, n)≤2n. The question is:
Question 2 . Give the exact value of C(d, n)for 2≤d≤n. If this is not possible, give lower and upper bounds as accurate as possible.
A well-known duality yields another formulation of Question 2. Let ϕ : S → 2F, a 7→ {E ∈ F ; a∈E} and setS =ϕ(S). In this manner, we have for alla∈S and allE ∈ F:
a ∈E ⇔E ∈ϕ(a). (1)
One can check that F shatters A ⊆ S if and only if, for every partition (B, C) of A (i.e.
A= B∪C and B ∩C =∅) the intersection T
b∈Bϕ(b)
∩ T
c∈Cϕ(c)
is nonempty, where the notationY stands for F \Y.
If IndF ≥ 2, then ϕ is a one-to-one correspondance from S to S, hence we have logn ≤ C(d, n) for all 2 ≤d ≤n, where log denotes the logarithm in base 2.
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The case d = 2
. Using for instance the binary expansion, it is easy to show that the order of magnitude ofC(2, n) is actually logn. The next statement refines this.Proposition 2 . Ifn = 12 2ll
= 2ll−−11
, then C(2, n) = 2l.
Proof. (Recall the notation A=F \A.) We first prove by contradiction that C(2, n)>2l−1.
Actually, if a family F of subsets of S shatters all 2-subsets of S, then the image S ⊆2F of S by ϕ must satisfy
∀A6=B ∈ S, A∩B, A∩B, A∩B, and A∩B are nonempty. (2) In particular S is a Sperner family of F (i.e. an antichain for the partial order of inclusion;
one finds several other expressions in the literature: ‘Sperner system’, ‘independent system’,
‘clutter’, ‘completely separating system’, etc.). For a survey on Sperner families and several generalizations, we refer e.g. to [1] and the references therein.
Assume now that |F | = 2l−1; it is known [6, 2, 3] that all Sperner families of F have a cardinal at most 2ll−−11
, and that there are only two Sperner families of maximal cardinal: the families l−F1
and Fl
, i.e. of (l−1)-subsets, resp. l-subsets of F. However, none of these families satisfies both A∩B and A∩B nonempty in (2). As a consequence, we must have
|F | ≥2l.
Conversely, let S = {a1, . . . , an}, consider {1,...,2ll }
, the set of l-subsets of {1, . . . ,2l}, and choose one element in each pair of complementary l-subsets. We then obtain a family {A1, . . . , An} which satisfies (2). Now we setF ={E1, . . . , E2l}, withEi ={aj ; i∈Aj}. The characterization (1) shows that F shatters every 2-subset of S.
The proof of the following statement is straightforward.
Corollary 3. If 2ll−−11
< n≤ 2l+1l
, then2l ≤C(2, n)≤2l+ 2.
The upper bound can be slightly improved: One can prove that, if 2ll−−11
< n ≤ l−2l1 , then 2l≤C(2, n)≤2l+ 1.
Question 3 . It seems that we have C(2, n) = k if and only if ⌊(k−k1)/2−2⌋−1
< n ≤ ⌊k/2k−⌋−1 1 , where ⌊x⌋ denotes the integer part of x. Is it true? Is it already known?
The first values are C(2,2) = C(2,3) = 4, C(2,4) = 5, C(2,5) = · · · = C(2,10) = 6. Com- puter seems to be useless, at least for a naive treatment. Already in order to obtainC(2,11) = 7, we would have to verify that C(2,11) > 6, i.e. to find, for each of the 2116
≈1017 families F in 2S some 2-subset that is not shattered by the family. (Alternatively, in the dual statement, we have to check “only” 2116
≈7.1011 families S in 2F.)
The case d ≥ 3
. From now, we assumen ≥4.Proposition 4 . For all3≤d < n, we have C(d, n)≤ 2d!d (3 logn)d.
The constant 3 can be improved. The proof below shows that, for all a > 1 and all n large enough, C(d, n)≤ 2d!d(alogn)d.
Proof. Let F0 ⊂ 2S be a minimal separating system of S, i.e. such that, for all a, b∈ S there exists Eab ∈ F0 which satisfies b /∈ Eab ∋ a. Since this amounts to choosing F0 minimal such that S = ϕ(S) is a Sperner family for F0, we know that |F0| =N if and only if ⌊(NN−−1)/21 ⌋
<
n ≤ ⌊N/2N ⌋
, hence N := |F0| ≤ 2 + logn + 12log logn ≤ 3 logn since n ≥ 4. We assume 2
N ≥ 2 in the sequel. Given two disjoint subsets B and C of S such that |B ∪C|=d, the set EBC = T
c∈C
S
b∈BEbc
contains B and does not meet C. Let F be the collection of all such sets EBC; then F shatters all subsets of S of cardinal at most d.
To estimate |F |, we consider Fk the collection of all such sets EBC, with |B| =k (and thus
|C|=d−k). We have |Fk| = Nk N−k
d−k
(with N =|F0|). Then we choose F =Sd
k=0Fk. We obtain|F | ≤Pd
k=0 N
k
N−k
d−k
= Nd
2d ≤ 2d!dNd≤ 2d!d (3 logn)d.
Question 4 . Is (logn)⌊d/2⌋⌊(d+1)/2⌋ the right order of magnitude for C(d, n)?
By constructing auxiliary Sperner families fromS, it is possible to give a better lower bound for C(d, n) than onlyC(d, n)≥C(2, n). For instance, in the cased= 3, for all distinctA, B, C ∈ S, we must haveA∩B 6⊆C. One can check that this implies that the family{A∩B ; A, B ∈ S}
is a Sperner family, therefore we obtain n2
≤ ⌊C(3,n)C(3,n)⌋/2
. Unfortunately, this does not modify the order of magnitude. Already in this cased= 3, we do not know whetherC(3, n) is of order logn, (logn)2, or an intermediate order of magnitude. Another formulation is:
Question 5 . Prove or disprove: There exists C > 0 such that, for all k ∈N, if F is a finite set of cardinalk and S ⊆2F satisfies ∀A, B, C ∈ S, A∩B 6⊆C, then|S| ≤C2C√k.
References
[1] P. Borg, Intersecting families of sets and permutations: a survey. Int. J. Math. Game Theory Algebra 21 (2012) 543–559.
[2] G. Katona, On a conjecture of Erd¨os and a stronger form of Sperner’s theorem. Studia Sci. Math. Hungar. 1 (1966) 59–63.
[3] D. Lubell, A short proof of Sperner’s theorem, J. Combin Theory 1 (1966) 299.
[4] N. Sauer, On the density of families of sets, J. Combin. Theory25 (1972) 80–83.
[5] S. Shelah, A combinatorial problem, stability and order for models and theories in infinite languages, Pacific J. Math. 41 (1972) 247–261.
[6] E. Sperner, Ein Satz ¨uber Untermenger einer endlichen Menge,Math. Zeitschrift 27 (1928) 544–548.
[7] V. N. Vapnik and A. Y. Chervonenkis, On the uniform convergence of relative frequences of events to their probabilities Theory Probab. Appl. 16 (1971) 264–280.
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